# Find all non-trivial submodules of a direct sum of two non-isomorphic simple modules

Let $R$ be a ring with $1$. Let $M_1$ and $M_2$ be two non-isomorphic simple (nonzero) $R$-modules. Find all non-trivial submodules of $M_1 \bigoplus M_2$.

Solution: $M_1 \bigoplus M_2 \cong M_1 \times M_2$. The submodules of $M_1 \times M_2$ are $M_1 \times \{0\}$, $\{0\} \times M_2$, $M_1 \times M_2$ and $\{0\} \times \{0\}$

• Your solution is technically correct, but it is also very incomplete. It includes some unnecessary things (like the $N$ and the $K$), and does not anywhere mention why the fact that the two modules are not isomorphic is important (this description of the submodules would be wrong otherwise). Apr 20, 2015 at 8:28
• @TobiasKildetoft if they are isomorphic then, we only have $M_1 \times M_2$ and $0 \times 0$. But I don't know how to explain it. I normally approach this question by trial and error. I don't really know how to do it sytematically.
– 123
Apr 20, 2015 at 13:04
• No, if they are isomorphic then we have at least one extra submodule (the diagonal), possible more. Apr 20, 2015 at 13:04
• @TobiasKildetoft what do you mean by the diagonal?
– 123
Apr 20, 2015 at 13:05
• I mean the subset consisting of all elements of the form $(x,x)$. Apr 20, 2015 at 13:07

Here's how you prove there aren't any submodules besides the obvious ones listed in the question. Suppose $N\subseteq M_1\oplus M_2$ is a submodule. Consider the projection $p:N\to M_1$. If the image of $p$ is $0$, then $N$ is contained in $0\oplus M_2\cong M_2$, so since $M_2$ is simple the only possibilities for $N$ are then $0\oplus 0$ and $0\oplus M_2$.

Since $M_1$ is simple, the only other possibility is that $p$ is surjective. In that case, consider the kernel of $p$, which is contained in $0\oplus M_2\cong M_2$. If the kernel of $p$ is $0$, then $p$ is an isomorphism. Now consider the projection $q:N\to M_2$. Since $N\cong M_1$ and $M_1$ and $M_2$ are non-isomorphic simple modules, $q=0$. The argument of the previous paragraph now shows that $N\cong 0\oplus 0$ or $M_1\oplus 0$ (actually it must be the latter since we are assuming $p$ is surjective).

The final case is that $p$ is surjective and the kernel of $p$ is $0\oplus M_2$. In that case $0\oplus M_2\subseteq N$ but $p$ is surjective on $N$, which implies $N$ is all of $M_1\oplus M_2$ (surjectivity means elements of $N$ can have any first coordinate, and then you can change the second coordinate arbitrarily).

• Very nice. Bounty awarded. Dec 19, 2016 at 3:14

Consider the composition series $$0\subset M_1\subset M_1\oplus M_2$$

If $N$ is a non-trivial submodule, then the series $$0\subset N\subset M_1\oplus M_2$$ can be refined to a composition series.

Since composition series have the same length, that means $0\subset N\subset M_1\oplus M_2$ is already a composition series, which is equivalent to the first by Jordan-Holder Theorem.

That is, $N\cong M_1$ or $M_2$.

Currently, that is what I can think of. I would be interested in a simpler solution to this question.

• Your argument doesn't distinguish b/w the case when $M_1$ and $M_2$ are isomorphic or not, b/c you haven't determined all the actual possible submodules, but just determined the possible isomorphism classes. Dec 13, 2016 at 17:48
• To get a complete classification, consider the morphisms $N \to M_i$ given by projecting on the $i = 1$ and $i = 2$ factors, and (assuming now that $M_1$ and $M_2$ are not isomorphic) see what you can deduce. Dec 13, 2016 at 17:49
• @tracing I can deduce that the projections are submodules of $M_1$ and $M_2$ respectively, and thus can only be 0 or $M_i$. However I am having some problems proving $N$ is the direct sum of the projections. Dec 13, 2016 at 23:48
• You know from your argument with composition series that either $N$ equals $M_1\oplus M_2$, or that $M = 0$, or (let's refer to this as the non-trivial case) that $M$ is isomorphic to one of $M_1$ or $M_2$. In particular, $N$ is simple in the interesting case. You have to use this ... . Dec 13, 2016 at 23:51
• More or less ... there is not really a canonical map $\pi_1(N) \to N$, so I would phrase it a little differently: e.g.: $\pi_i$ restricted to $N$ is either an isomorphism or zero, and (since $M_1$ and $M_2$ are not isomorphic) both can't be isomorphisms, and (since $N$ embeds into $\pi_1(N)\times \pi_2(N)$) both can't be zero. Thus exactly one of them is an iso., while the other is zero, so that either $N = M_1\times 0$ or $N = 0 \times M_2$. Dec 14, 2016 at 1:09